Answer:
So as sunlight continued to enter the box and the walls continued to warm, heat built up inside and increased the temperature.
What will be the current through a resistance of 50Ω if the applied voltage across the resistance is 117V?
Formulas: V=IR, I=V/R, R=V/I
Note: No need to write the unit of your answer.
2.34 A
Explanation:
[tex]V = IR \Rightarrow I = \dfrac{V}{R} = \dfrac{117\:\text{V}}{50\:Ω} = 2.34\:\text{A}[/tex]
An astronaut standing on a platform on a foreign planet drops a hammer. If the hammer falls 9.0 meters vertically in 2.5 seconds, what is the acceleration due to gravity on that planet?
Answer: 1.646 m/s²
Explanation:
The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,
d = Vot + 0.5at²
where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,
6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²
Determining the value of a,
a = 1.646 m/s²
Find the angle between two equal forces (p) when their resultant is equal to (p)
Answer:
60°
Explanation:
It will be equilateral triangle, you can prove with the cosine theorem.
1)P^2=P^2+P^2 - 2×P×P× cos A
2) cos A= P^2 / 2× P^2 = 1/2
cos A = 1/2
A = 60°
If you wrote this question right,it will be solved like this :)
Technician a says that the piston engine starts and stops several times per 2nd technician B says that with the rotary engine each rotor has 3 working chambers so it acts like a 3 cylinder engine while it is at operating
Answer:
Explanation:
trfbehherebgqerfherbfhyqerwbfvhrbhferhgvr
1. A small package is dropped from the Golden Gate Bridge. What is the velocity of the package
after it has fallen for 3.0 3?
The velocity of the package after it has fallen for 3.0 s is 29.4 m/s
From the question,
A small package is dropped from the Golden Gate Bridge.
This means the initial velocity of the package is 0 m/s.
We are to calculate the velocity of the package after it has fallen for 3.0 s.
From one of the equations of kinematics for objects falling freely,
We have that,
v = u + gt
Where
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
and t is time
To calculate the velocity of the package after it has fallen for 3.0 s
That means, we will determine the value of v, at time t = 3.0 s
The parameters are
u = 0 m/s
g = 9.8 m/s²
t = 3.0 s
Putting these values into the equation
v = u + gt
We get
v = 0 + (9.8×3.0)
v = 0 + 29.4
v = 29.4 m/s
Hence, the velocity of the package after it has fallen for 3.0 s is 29.4 m/s
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What happens to the size of an Object when it is heated ?
Answer:
the object melts so the size get smaller
The surface of a running track is made of rubber. A heavy trolley is pulled on to the track and it exerts a large force on the rubber track. State two effects that this force has on the rubber
Answer:
hi im from canada so can we be friends
Select the correct answer.
The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's motion?
OA
It's not moving
ОВ.
It's moving at a constant speed.
OC. It's moving at a constant velocity.
OD. It's speeding up.
Reset
Next
Answer:
A.It’s not moving.
Explanation:
Position-Time graphs display the motion of a object by showing the changes of velocity with respect to time.
The motion of a car on a position-time graph that is represented with a horizontal line indicates that the car has stopped moving.
A straight line with a positive slope indicates that the car is moving at a constant velocity, and thus the slope is constant. On the other hand, a curve with a changing slope, shows that the velocity is changing.
The pulley shown in the attached diagram has a diameter of 30 centimeters and a mass of 19 kilograms. The pulley is a solid disk with an axle through its center.
(a.) What is the moment of inertia of the pulley?
(b.) What is the magnitude of the net torque on the pulley about its axis
(c.) What is the direction of the net torque on the pulley?
(d.) What is the magnitude of the angular acceleration of the pulley?
(e.) What is direction of the angular acceleration?
Answer:
Explanation:
a) I = ½mR² = ½(19)(0.15²) = 0.21375 kg•m²
b) τ = Fnet(r) = (25 - 12)(0.15) = 1.95 N•m
c) CCW
d) a = τ/I = 1.95 / 0.21375 = 9.12280701... = 9.1 rad/s²
e) CCW
You are observing a star that is 250 million light years away. From the starlight you are observing now, the appears to be 30 million years old and you determine that the star has a lifetime of 120 million years. How long will it be until Earth receives light from the supernova that occurs at the end of the life of the star
What ever games your playing it wont work you cant defeat me hahhahha oh I know :O but he can AHHHH :D D: NOOOO IMPOSSIBLE.
PLZ HURRY ANSWER THIS PLZ
Which statement best describes how Greensburg recovered from the tornado? O A. It doubled its population O B. It secured itself against future tornadoes. C. It became an eco-tourism destination © D. It built a new school and hospital.
Answer:
C: it became an eco-tourism destination
Explanation:
Answer: Its C
Explanation:
Someone help please i need to finish this
Answer:
4th answer
Explanation:
The gradient of a distance-time graph gives the speed.
gradient = distance / time = speed
Here, the gradient is a constant till 30s. So it has travelled at a constant speed. It means it had not accelarated till 30s. and has stopped moving at 30s.
Ano ang mangyayari kung hindi mo gagamitin ang iyong talento?
Answer:
mas lalo kang tinatamaran magaral
1.protozoo ore ?
A. multicellular organisms B. one - Celled animals
C. amembers of the group Protocista
D. unicellulor Plants
Answer:
c
Explanation:
amembers of the group protocista
if two masses 5.2kg and 4.8kg are attached to ends of inextensible string passed over a friction less pulley then acceleration of system
Answer:
If two masses 5.2 kg and 4.8 kg are attached to the ends of inextensible strings passed over a frictionless pulley then the acceleration of system is 0.4 ms-2.
Explanation:
what is one measurement needed to calculate the speed of an object
Answer:
The equation for speed is simple: distance divided by time. You take the distance traveled (for example 3 meters), and divide it by the time (three seconds) to get the speed (one meter per second).
Explanation:
25. Which of the following cannot be broken down into smaller parts through ordinary chemical means?
a. Nitrogen
b. Protein
C. Salt
d. Sugar
Answer:
the correct answer is sugar
HELP!!! I HAVE NO IDEA!!!!!!!!!!!!!
1. A rocket has a mass of 0.8 kg and an engine that provides 100 N of force. A second rocket is being designed to use the same engine but accelerate at half the rate of the first rocket. What is the mass of the second rocket?
a. 0.4 kg
b. 1.6 kg
c. 2.4 kg
d. 0.8 kg
2. Which of the following would be a situation with unbalanced forces?
a. Two people pulling on the same side of a wheelbarrow
b. A team of players in tug of war pulling on the rope, each team with equal numbers of people pulling with equal strength
c. Two people of opposite sides of a big tire. One pushes the tire and one pulls it with equal force
d. Two people not touching a crate that is sitting stationary.
3. Why is it generally easier to push a heavy object that is already moving instead of the same heavy object that is stationary
a. The force of gravity on the object is different if it is stationary or not.
b. The force of friction on the object is different if it is stationary or not
c. The force of the push on the object is different if it is stationary or not
d. The normal force on the object is different if it is stationary or not.
Thanks In Advance!! (I will mark brainliest) :)
Answer:
Explanation:
The engine should create the same force
1) F = ma
F = M(½a)
therefore
ma = ½Ma
m = ½M
M = 2m
M = 2(0.8)
M = 1.6 kg
2) Unbalanced forces
a. Two people pulling on the same side of a wheelbarrow
c. Two people of opposite sides of a big tire. One pushes the tire and one pulls it with equal force
3) b. The force of friction on the object is different if it is stationary or not
When converted to a household measurement, 9 kilograms is approximately equal to a
Answer:
D) 19.8 lbs
Explanation:
1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.
1kg is also equal to 2.205 lbs. 9*2.205=19.8416
9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system
a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.
This leaves us with 19.8 lbs
Which of the following statements is true?
a Penicillin is produced by fungal fermentation.
b
The effects of too much beer or wine can be
countered with antibiotics.
с
The deadly Death Cap mushroom can be treated
with penicillin.
d Antibiotics do not treat fungal diseases.
Answer:
a Penicillin is produced by fungal fermentation.
Explanation:
The industrial production of penicillin (PEN) occurs via fermentation using the filamentous fungus Penicillium chrysogenum
Penicillin G (benzylpenicillin) was first produced from a penicillium fungus that occurs in nature. The strain of fungus used today for the manufacture of penicillin G was created by genetic engineering to improve the yield in the manufacturing process.
In hockey activities, a warm hockey puck and a frozen hockey puck has a different coefficient of restitution: 0.5 for a warm hockey puck, and 0.35 for a frozen one. NHL requires the frozen pucks to be used in games. To make sure the puck can be used in the game, the referee drops the puck on its side from a height of 2.5 m. How high should the puck bounce if it is a frozen puck
If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
Given the data in the question;
Since the hockey puck was initially in the referee's hands
Initial velocity; [tex]u = 0m/s[/tex]Distance or height from which it was dropped; [tex]h = 2.5m[/tex]Acceleration due to gravity; [tex]g = 9.8 m/s^2[/tex]Coefficient of restitution a frozen puck; [tex]0.35[/tex]First we will find the velocity of the Puck when it hits the ground
From the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
[tex]v^2 = u^2 + 2gh[/tex]
We substitute in our value and find "v"
[tex]v^2 = 0 + (2 \ *\ 9.8m/s^2\ *\ 2.5m )\\\\v^2 = 47.04m^2/s^2\\\\v= \sqrt{47.04m^2/s^2}\\\\v = 6.85857m/s[/tex]
Now, Velocity of the hock puck after it hits the ground and bounce back;
We know that; Coefficient of restitution [tex]= \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}[/tex]
Hence, Relative Velocity after collision = Coefficient of restitution × Relative Velocity before collision
we substitute in our values;
Relative Velocity after collision [tex]= 0.35 \ *\ 6.85857m/s[/tex]
Relative Velocity after collision [tex]= 2.4 m/s[/tex]
Now, to determine how high should the puck bounced back
We use the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
[tex]v^2 = u^2 + 2gh[/tex]
Now, since the hockey puck bounces back, it is experiencing a negative acceleration
Hence, the equation becomes
[tex]v^2 = u^2 - 2gh[/tex]
We substitute our values into the equation and find "h"
[tex](0m/s)^2 = (2.4m/s)^2 - ( 2*9.8m/s^2*h)\\\\0 = 5.76m^2/s^2 - (19.6m/s^2*h)\\\\(19.6m/s*h) = 5.76m^2/s^2 \\\\h= \frac{ 5.76m^2/s^2 }{19.6m/s^2}\\\\h = 0.3m[/tex]
Therefore, If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
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Jasmine pushing her little brother on the swing the swing passes back-and-forth 14 times in 40 seconds what is the frequency of the motion and what is superior to the swing
Answer:
Explanation:
frequency f = 14 cycles/40 s
f = 0.35 cycles/s or
f = 0.35 Hz.
I suspect you are more concerned with the PERIOD of the swing rather than what is superior to it. (How would one even compare?)
T = 1/f = 1/0.35
T = 2.85714285...
T = 2.86 s
Verificar
El gráfico de la velocidad de una bola que rueda por un canal se muestra en la
figura:
a) La velocidad de la bola luego de 5 s.
0.6m/s
b) La distancia recorrida en el intervalo t : (0-5) s
1.5m
c) La distancia recorrida en el intervalo t: (5-10) s
3m
d) La aceleración de la bola en el intervalo t: (10-12) s.
-0.4m/s^2
e) La distancia total recorrida.
6.9m
f) La rapidez media de la bola
0.58m/s
El análisis de un grafico de velocidad versus tiempo permite encontrar los resultados para las preguntas son:
a) La velocidad es v= 0,6 m/s
b) La distancia recorrida dₐ = 1,5 m
c) La distancia recorrida entre t= 5 y 10 s es [tex]d_b[/tex] = 3,0 m
d) La aceleración a = -0,4 m/s²
e) La distancia total recorrida es [tex]d_{total}[/tex] = 5,3 m
f) La velocidad media es [tex]v_{media}[/tex] = 0,44 m/s
La cinemática estudia el movimiento de los cuerpos, buscando relaciones entre la posición, la velocidad y la aceleración de los cuerpos.
La velocidad es definida como el cambio de la posición en función del tiempo y la aceleración es el cambio de la velocidad en el intervalo de tiempo.
.[tex]v= \frac{\Delta x}{t} \\a = \frac{\Delta v}{t}[/tex]
Donde v y a son la velocidad y aceleración, respectivamente, t es el tiempo y Δx y Δv son la variación de la posición y la velocidad, respectivamente.
En el adjunto muestran un grafico de la velocidad en función del tiempo, donde del área bajo la curva obtenemos la distancia recorrida y la pendiente corresponde a la aceleración del cuerpo.
Respondamos las preguntas:
a) La velocidad de la bola a los 5 s.
Como es un grafico de velocidad versus tiempo, la velocidad puede ser leída directamente, en el grafico tenemos un valor de
v= 0,6 m/s
b) La distancia recorrida corresponde al área bajo la curva.
[tex]d_a = \frac{v_f - v_o}{2} t \\d_a = \frac{0.6-0}{ 2}\ 5[/tex]
d = 1,5 m
c) La distancia en el intervalo de 5 a 10 s.
En este intervalo la velocidad es constante, el área es
[tex]d_b = v \Delta t\\d_b = 0.6 \ 5\\[/tex]
d = 3 m
d) La aceleración en el intervalo 10 a 12 s.
La aceleracion es la pendiente del grafico
a = [tex]\frac{\Delta v}{\Delta t}[/tex]
a = [tex]\frac{-0.2 - 0.6}{12-10}[/tex]
a = -0,4 m/s²
El signo negativo indica que la aceleración se opone a la velocidad del cuerpo.
e) La distancia total recorrida
[tex]d_{total} = d_a + d_b + d_c[/tex]
dₐ = 1,5 m
[tex]d_b[/tex] = 3,0 m
Busquemos la distancia en el ultimo intervalo
[tex]d_c = \frac{0.6 + 0.2}{2} ( 12-10)[/tex]
[tex]d_c[/tex] = 0,8 m
Substituimos
[tex]d_{total}[/tex] = 1,5 + 3,0 + 0,8
[tex]d_{total}[/tex] = 5,3 m
f) La rapidez de la bola
La rapidez prmedio es la relación entre la distancia recorrida en el tiempo.
[tex]v_{avg} = \frac{\Delta x}{\Delta t }[/tex]
[tex]v_{avg} = \frac{5.3}{12}[/tex]
[tex]v_{avg}[/tex] =0,44 m/s
En conclusión usando el análisis de un grafico de velocidad versus tiempo podemos encontrar los resultados para las preguntas son:
a) La velocidad es v= 0,6 m/s
b) La distancia recorrida da = 1,5 m
c) La distancia recorrida entre t= 5 y 10 s es db = 3,0 m
d) La aceleración a = -0,4 m/s²
e) La distancia total recorrida es [tex]d_{total}[/tex] = 5,3 m
f) La velocidad media es [tex]v_{avg}[/tex] = 0,44 m/s
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A 500 kg car is at rest at the top of a 72 m high hill. The car rolls to the bottom of the hill. At the bottom of the hill, the car has a speed of 25.6 m/s. Calculate the mechanical energy of the car at the top and bottom of the hill. (Assume the bottom of the hill has a height of 0 m, g=9.80 ms2/).
Explanation: Solution
1.
Gravitational potential energy
U=mgh=500*9.8*50
U=245000 J
2.
Kinetic energy is present at bottom of the hill
K=(1/2)mV2=(1/2)*500*27.82
K=193210 J
3.
Work done by friction
W=193210-245000=-51790 J
The mechanical energy at the top and bottom of the hill is equal to 352800 J and 163840 J respectively.
What is the kinetic energy and potential energy?Kinetic energy (KE) can be described as the energy possessed by a moving object due to its motion. Work by a body will be done to change the kinetic energy. The kinetic energy is represented as K.E = ½mv².
Potential energy (P.E) can be described as the energy that is stored by an object due to its position and is represented in the equation as P.E = mgh, where ‘m’ is the mass, ‘g’ is the acceleration due to gravity and ‘h’ is the height.
The mechanical energy = Kinetic energy + potential energy
Given, the mass of the car, m = 500 Kg
The height of the hill, h = 72 m
The velocity of the car, v = 25.6 m/s
At the top of the hill, the mechanical energy = potential energy
The potential energy at the top of the hill, = mgh
P. E. = 500 × 9.8 ×72
P.E. = 352800 J
At the bottom of the hill, the mechanical energy = kinetic energy
The kinetic energy of the car at the bottom of the hill,
K.E. = ½ × 500 (25.6)²
K.E. = 163840 J
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Indicate whether each of the following statements about car collisions is true or false. Select all that are True.
a. If two identical cars with identical speeds collide head on, the magnitude of the impulse received by each car and each driver is the same as if one car at the same speed had collided head on with a concrete wall.
b. Car 1 has mass m, and car 2 has mass 2m. In a head-on collision of these cars while moving at identical speeds in opposite directions, car 1 experiences a bigger acceleration than car 2.
c. The essential safety benefit of crumple zones (parts of the front of a car designed to receive maximum deformation during a head-on collision) is due to their absorbing kinetic energy, converting it into deformation, and lengthening the effective collision time, thus reducing the average force experienced by the driver.
d. Car 1 has mass m, and car 2 has mass 2m. In a head-on collision of these cars while moving at identical speeds in opposite directions, car 1 receives an impulse of bigger magnitude than that received by car 2. If car 1 has mass m and speed v, and car 2 has mass 0.5m and speed 1.5v, then both cars have the same momentum.
Answer:
True statements a, b, and c
Explanation:
Answer d is false because in any collision, each object receives the same impulse. Also the momentum of car 1 is mv, while the momentum of car 2 is 0.5m(1.5v) = 0.75mv
The impulse of collision of cars depends on the mass and velocity of the cars. The statements which are true about the car collisions are option a, b, c.
What is impulse?Impulse is a physical quantity expressed as the product of force and time. The change in momentum mΔV is numerically equals to the impulse F t.
If two identical cars with identical speeds collide head on, the magnitude of the impulse received by each car and each driver is the same as if one car at the same speed had collided head on with a concrete wall. Because they have equal mass and velocity.
The bigger acceleration is experienced by the one which have lighter mass thus car 1 with lower mass experience higher acceleration .
The essential safety benefit of crumple zones (parts of the front of a car designed to receive maximum deformation during a head-on collision) is due to their absorbing kinetic energy.
Converting this energy into deformation, and lengthening the effective collision time, thus reducing the average force experienced by the driver.
The momentum change is proportional to the velocity. The one which have higher velocity is having the higher momentum. Thus car 2 have higher velocity. Thus option d is false.
Therefore, the statements which are true are a, b and c.
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A spring of spring constant k = 200 N m−1 is slowly extended from an extension of 3.0 cm to an extension of 5.0 cm. Calculate the work done by the extending force. 60
Answer:
31
Explanation:
No need
A water rocket uses an amount of water and pressurized air to send a plastic rocket several feet into the air. As the water and air rush out the tail end of the rocket, the rocket shoots into the air. Which statement explains the rocket's motion?
Answer:
D.
For every action there is an equal and opposite reaction.
Explanation:
im doing the same one lol
For every action there is an equal and opposite reaction explains the rocket's motion.
What is rocket?A rocket is indeed a vehicle that accelerates by employing jet propulsion rather than the surrounding air. A rocket engine generates thrust by reacting to high-speed exhaust. Since rocket engines run solely on fuel carried within the vehicle, a rocket may travel in space.
A water rocket uses an amount of water and pressurized air to send a plastic rocket several feet into the air. As the water and air rush out the tail end of the rocket, the rocket shoots into the air. For every action there is an equal and opposite reaction explains the rocket's motion.
Therefore, for every action there is an equal and opposite reaction explains the rocket's motion.
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How can stretching affect the range of motion of the neck? Hypothesis
Answer:
reduce passive stiffness and increase range of movement during exercise.
Explanation:
stretching performed as part of a warm up prior to exercise is thought to reduce passive stiffness and increase range of movement during exercise. in general it appears that is static stretching is most beneficial for athletes requiring flexibility for their sports.
HELPPPP
The maximum force of sliding friction between a 10 kg rubber box and the concrete
floor is 64 N. How much force should a worker push on the box with if he wants it to
move at a constant velocity?
1) A little less than 64 N
2)A little more than 64 N
3)Exactly 64 N.
4)Exactly 640 N
The force that will move the box at constant velocity must be a little more than 64 N.
The coefficient of sliding friction is obtained from the formula;
μ= F/R
Where;
F = frictional force
μ = coefficient of sliding friction
R = Normal reaction
It is necessary to note that the force that will move the body must be greater than the frictional force acting between the body and the surface in order to move the body. Hence, the force that will move the box at constant velocity must be a little more than 64 N.
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A faulty fireworks rocket launches but never discharges. If the rocket launches
with an initial velocity of 33.0 ft/s at an angle of 85.0°, how far away from the
launch site does the rocket land?
Answer:
a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0∘ above the horizontal. During the flight, the rocket explodes into two pieces of equal mass (see Fig. 8.32). (a) What horizontal distance from the launch point will the center of mass of the two pieces be after both have landed on the ground? (b) If one piece lands a horizontal distance of 26.0 m from the launch point, where does the other piece land?
